Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $p = \dfrac{-3n^2 - 27n}{-3n^3 - 12n^2 + 135n} \div \dfrac{n + 8}{-3n^2 + 3n + 216} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{-3n^2 - 27n}{-3n^3 - 12n^2 + 135n} \times \dfrac{-3n^2 + 3n + 216}{n + 8} $ First factor out any common factors. $p = \dfrac{-3n(n + 9)}{-3n(n^2 + 4n - 45)} \times \dfrac{-3(n^2 - n - 72)}{n + 8} $ Then factor the quadratic expressions. $p = \dfrac {-3n(n + 9)} {-3n(n + 9)(n - 5)} \times \dfrac {-3(n + 8)(n - 9)} {n + 8} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac {-3n(n + 9) \times -3(n + 8)(n - 9) } { -3n(n + 9)(n - 5) \times (n + 8)} $ $p = \dfrac {9n(n + 8)(n - 9)(n + 9)} {-3n(n + 9)(n - 5)(n + 8)} $ Notice that $(n + 9)$ and $(n + 8)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {9n(n + 8)(n - 9)\cancel{(n + 9)}} {-3n\cancel{(n + 9)}(n - 5)(n + 8)} $ We are dividing by $n + 9$ , so $n + 9 \neq 0$ Therefore, $n \neq -9$ $p = \dfrac {9n\cancel{(n + 8)}(n - 9)\cancel{(n + 9)}} {-3n\cancel{(n + 9)}(n - 5)\cancel{(n + 8)}} $ We are dividing by $n + 8$ , so $n + 8 \neq 0$ Therefore, $n \neq -8$ $p = \dfrac {9n(n - 9)} {-3n(n - 5)} $ $ p = \dfrac{-3(n - 9)}{n - 5}; n \neq -9; n \neq -8 $